The solubility product of PbI2 is 7.2×10−9. Calculate the maximum m – InterviewSolution

1.	The solubility product of PbI2 is 7.2×10−9. Calculate the maximum mass of NaI, which may be added in 500 mL of 0.005 M Pb(NO3)2 solution without any precipitation of PbI2(Atomic weight of Na=23 u, I=127 u)
Answer» The solubility product of $P b I_{2}$ is $7.2 \times 10^{- 9}$ . Calculate the maximum mass of $N a I$ , which may be added in $500 mL$ of $0.005 M$ $P b (N O_{3})_{2}$ solution without any precipitation of $P b I_{2}$ (Atomic weight of $N a = 23 u, I = 127 u$ )

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